You posted a question about having two cannon balls strike the same target simultaneously, with a time delay between shots:

https://www.jiskha.com/display.cgi?id=1520556373

I have had some time to think about it, and have come up with the following ideas. Maybe you have solved it in the meantime.

Since the range of a shot fired with velocity v at angle θ is
R(θ) = v^2/g sin2θ
then if the 2nd shot, fired at angle Ø with the same speed is the same, we need
sin2θ = sin2Ø
That means that 2θ = 180-2Ø, or Ø=90-θ

Our cannonball was fired with v=1000, and the second shot was delayed k seconds.
Since the time in the air at angle θ is 2v/g sinθ = 200sinθ, we need to maximize the distance traveled by the ball. That is, maximize the flight time. This is clearly when θ=90 degrees.

But, that means the ball goes straight up, making it impossible to fire a 2nd ball at a lower angle and hit the same spot. So, pi/4 < θ < pi/2.

So, with a delay of k seconds, we have

200sinθ = 200sinØ + k
k = 200(sinθ-sinØ)
But since Ø=90-θ, sinØ=cosθ and so
k = 200(sinθ-cosθ)

Sanity check: If the time delay is zero, then sinθ=cosθ and θ=45. This provides the maximum range, so no time delay can be allowed. But, we need k>=60

Since we need to wait at least a minute, k >= 60, meaning that
200(sinθ-cosθ) > 60
sinθ-cosθ >= 0.3

Now we know that we need to maximize θ subject to the constraint that
pi/4 < θ
sinθ-cosθ > 0.3

This means that 0.999154 < θ < pi/2
In other words, we basically need θ=1

check:
Range(θ=1) = 90929 -- k=60.24
Range(θ=1.1) = 80849 -- k=87.52
Range(θ=0.9) = 97384 but the time lag is only 32.37 seconds

So, you can see that larger values of θ result in a smaller range, and smaller θ means that you cannot wait the required 60 seconds before firing the 2nd shot.

It seems like you have made significant progress in solving the problem! However, let's go over the steps together to ensure we have the correct solution.

The problem is to find the time delay, denoted by k, between two cannonballs fired at the same target simultaneously. We assume both cannonballs have the same initial velocity, v=1000.

To address this problem, you started by using the equation for the range, R(θ), of a cannonball fired at an angle θ:

R(θ) = v^2/g sin(2θ)

You correctly recognized that for the range of the second shot to be the same as the first shot, we need sin(2θ) = sin(2Ø), where Ø is the angle of the second shot.

By applying the double angle formula, we can express Ø in terms of θ as Ø = 90 - θ. This substitution allows us to relate the angles of the two shots.

Next, you noted that to maximize the distance traveled by the ball, we need to maximize the flight time. Since the time in the air at angle θ is 2v/g sin(θ), we want to find the maximum range when the ball is in the air for the longest time.

Here you made a slight error in reasoning. The ball would achieve the longest flight time when it is shot straight up, which corresponds to θ = 90 degrees. However, as you correctly determined, firing a second ball at a lower angle in this case would not hit the same spot.

To address this issue, you correctly concluded that the first shot must have an angle within the range π/4 < θ < π/2. This range ensures that the ball is fired at an angle that allows time for the second shot to be fired with a delay of at least 60 seconds.

Using the expression for k = 200(sinθ - cosθ), you derived this equation by substituting sinØ = cosθ.

To perform a sanity check, you confirmed that when k = 0, sinθ = cosθ, meaning θ = 45 degrees, which is the angle that provides the maximum range. Consequently, k cannot be zero, as we need a minimum delay of 60 seconds.

Finally, by solving the inequality 200(sinθ - cosθ) > 60, you found that sinθ - cosθ ≥ 0.3. After some calculations, you obtained the range 0.999154 < θ < π/2, which essentially means that θ = 1 provides a suitable solution.

To verify this solution, you checked the ranges obtained for θ = 1, 1.1, and 0.9 and observed that the range is maximum for θ = 1 while satisfying the minimum delay requirement.

In summary, the optimal angle for the first shot is θ = 1, and the corresponding time delay for the second shot is approximately 60.24 seconds.